Question 50 Solve the problem. When obtaining a confidence...

Question 50 Solve the problem. When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population correction factor: sqrt((N-n)/(N-1)) Find the 95% confidence interval for the mean of 200 weights if a sample of 34 of those weights yields a mean of 155.7 lb and a standard deviation of 24.1 lb. A) 146.8 lb < ? < 164.6 lb B) 148.3 lb < ? < 163.1 lb C) 149.3 lb < ? < 162.1 lb D) 147.6 lb < ? < 163.8 lb

Transcript

00:02 Hello, in this video, we're going to find the 95 % confidence interval, which will denote as ci for a short -hand notation.

00:10 Given that the sample size is 34, the mean is 200 pounds, and the standard deviation is 24.

00:25 So now we need to find the confidence interval with 95 % confidence that the samples will fall inside this, that 95 % of the samples will fall inside this.

00:41 This interval.

00:42 So the formula for the confidence interval is the mean plus or minus the z score times the standard deviation divided by the square root of the number of the samples.

00:55 So the z score we can get from the table and for 95 % confidence interval our z score is equal to 1 .96.

01:09 So now we have everything we need to find this confidence interval.

01:13 Let's first look at the plus side so we're going to be using a plus and a subtraction so the upper bound of this conference interval will be up will be obtained by using the plus so let's plug in what we have you have the standard you have the mean as 200 pounds so 200 plus our z score of 1 .96 times the standard deviation which is 24 .1 pounds and then all of that is divided by the square root of the number of samples, which is 34.

02:00 So this will be our upper bound of the conference interval.

02:04 And similarly, we can do another calculation for the lower bound, which will basically use the same expression, but instead of a plus, we'll be using the minus.

02:17 So let's calculate this one first for the upper bound...

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